The parity of the pion from $\pi d\to n n$.

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We can determine the internal parity of the pion by studying pion capture by a deuteron, ${\pi} +{d}\rightarrow n+n.$ The pion is known to have spin 0, the deuteron spin 1, and the neutron spin ${1\over 2}$. The internal parity of the deuteron is +1. The pion is captured by the deuteron from a 1S states, implying $\ell=0$ in the initial state. So the total angular momentum quantum number of the initial state is $j=1$.

So the parity of the initial state is

$$\bgroup\color{black}(-1)^\ell P_\pi P_d = (-1)^0 P_\pi P_d = P_\pi\egroup$$

The parity of the final state is

$$\bgroup\color{black}P_nP_n (-1)^\ell=(-1)^\ell \egroup$$

Therefore,

$$\bgroup\color{black}P_\pi=(-1)^\ell.\egroup$$

Because the neutrons are identical fermions, the allowed states of two neutrons are $^1S_0$, $^3P_{0,1,2}$, $^1D_2$, $^3F_{2,3,4}...$ The only state with $j=1$ is the $^3P_{1}$ state, so $\ell=1$

$$\bgroup\color{black}\Rightarrow P_\pi =-1.\egroup$$